Codeforces Beta Round #5 C. Longest Regular Bracket Sequence

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經過了一個多月的時間，今天終於可以回到正軌了，繼續開始刷CF。

題目大意：

給出一個只有括弧的字串，求最長“匹配”子串的長度和數量。

解題思路：

設定數組記錄匹配括弧段的開頭。

下面是代碼：

#include <set>#include <map>#include <queue>#include <math.h>#include <vector>#include <string>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <cctype>#include <algorithm>#define eps 1e-10#define pi acos(-1.0)#define inf 107374182#define inf64 1152921504606846976#define lc l,m,tr<<1#define rc m + 1,r,tr<<1|1#define zero(a) fabs(a)<eps#define iabs(x)  ((x) > 0 ? (x) : -(x))#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (min(SIZE,sizeof(A))))#define clearall(A, X) memset(A, X, sizeof(A))#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))#define memcopyall(A, X) memcpy(A , X ,sizeof(X))#define max( x, y )  ( ((x) > (y)) ? (x) : (y) )#define min( x, y )  ( ((x) < (y)) ? (x) : (y) )using namespace std;char s[1000006];//char sta[1000006]int op[10000006];//int ed[10000006];int main(){    scanf("%s",s);    int top=0,len=strlen(s),bt=0;    int max1=0,maxcnt=1;    clearall(op,-1);    for(int i=0; i<len; i++)    {        if(s[i]=='(')        {            if(op[top]==-1)                op[top]=i;            top++;            // ed[top++]=i;        }        else        {            if(top==bt)            {                top++;                bt++;            }            else            {                if(i-op[top-1]+1>max1)                {                    max1=i-op[top-1]+1;                    maxcnt=1;                }                else if(i-op[top-1]+1==max1)                {                    maxcnt++;                }                op[top]=-1;                top--;            }        }    }    printf("%d %d",max1,maxcnt);    return 0;}

Codeforces Beta Round #5 C. Longest Regular Bracket Sequence